Edexcel S1 2002 January — Question 3 7 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2002
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeTwo unknowns from sum and expectation
DifficultyModerate -0.8 This is a straightforward S1 question testing basic probability distribution properties. Part (a) requires setting up E(X) = Σxp(x) and solving a simple linear equation. Part (b) uses the standard variance formula with values already found. Part (c) is trivial since X only takes values 0,1,2. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

A discrete random variable \(X\) has the probability function shown in the table below.
\(x\)012
P(\(X = x\))\(\frac{1}{3}\)\(a\)\(\frac{2}{3} - a\)
  1. Given that E(\(X\)) = \(\frac{2}{3}\), find \(a\). [3]
  2. Find the exact value of Var (\(X\)). [3]
  3. Find the exact value of P(\(X \leq 15\)). [1]

Part (a)
AnswerMarks Guidance
\(a + 2(\frac{3}{2} - a) = \frac{5}{6}\)M1 Use of \(E(X)\) correct equation
\(a = \frac{1}{3}\)A1
A1 (3)Allow \(\frac{1}{3}\)
Part (b)
AnswerMarks Guidance
\(\text{Var}(X) = [\frac{1}{3} \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} - (\frac{2}{3})^2]\)M1 Use of \(\sum x^2 P(x+x) - \mu^2\)
\(= \frac{17}{36} = 0.472\)M1 Allow \(\frac{17}{36}\)
A1 (3)
Part (c)
AnswerMarks
\(P(X \leq 1.5) = P(0) + P(1) = \frac{2}{3} + \frac{1}{2} = \frac{7}{6}\)B1(0)
## Part (a)
$a + 2(\frac{3}{2} - a) = \frac{5}{6}$ | M1 | Use of $E(X)$ correct equation
$a = \frac{1}{3}$ | A1 | 
 | A1 (3) | Allow $\frac{1}{3}$

## Part (b)
$\text{Var}(X) = [\frac{1}{3} \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} - (\frac{2}{3})^2]$ | M1 | Use of $\sum x^2 P(x+x) - \mu^2$
$= \frac{17}{36} = 0.472$ | M1 | Allow $\frac{17}{36}$
 | A1 (3) | 

## Part (c)
$P(X \leq 1.5) = P(0) + P(1) = \frac{2}{3} + \frac{1}{2} = \frac{7}{6}$ | B1(0) | 

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A discrete random variable $X$ has the probability function shown in the table below.

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
$x$ & 0 & 1 & 2 \\
\hline
P($X = x$) & $\frac{1}{3}$ & $a$ & $\frac{2}{3} - a$ \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}[label=(\alph*)]
\item Given that E($X$) = $\frac{2}{3}$, find $a$. [3]
\item Find the exact value of Var ($X$). [3]
\item Find the exact value of P($X \leq 15$). [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2002 Q3 [7]}}